# Probability of both kids being a girl given one is a girl

So I found this teaser:

A mom has two kids. Given one of them is a girl, what's the probability that both are girls?

The answer is 1/3. Which I have reasoned with the following.

There are four possible outcomes for two kids, BB, BG, GB, GG. Given that one is a girl, this limits us to three possible outcomes and excludes the BB possibility. One of those three outcomes is GG (assuming boys and girls are equally likely) giving us a probability of 1/3.

However I am having trouble reconciling this with an application of Baye's Thm. Lets say we articulate two events as A = Both are girls and B = one is a girl. Then we are looking for:

$$P(A|B)$$

Using Baye's

$$P(A|B) = \frac{P(B|A)*P(A)}{P(B)}$$

$$P(B|A) = 1$$, if both children are girls then obviously one of them is a girl. $$P(A) = 1/4$$ Assuming independence and a 50/50 chance. And $$P(B) = 1/2$$ Because both children have 50/50 chance of being a girl. However when I work through these numbers I get

$$P(A|B) = 1/2$$

Which is not the expect answer of 1/3 based on my logic above. I'm assuming something is wrong with my event descriptions but I'm not sure what

B should be the probability that AT LEAST one is a girl. Then $P(B) = 3/4$ and this math works out to the 1/3 answer